Answer
(a) $3.32Kg$
(b) $48.5\%$
(c) $87.1\%$
Work Step by Step
(a) The required mass of gold can be determined as:
$V_{gold}=\frac{m_{rock}-\rho_{granite}V_{rock}}{\rho_{gold}-\rho_{granite}}$
We plug in the known values to obtain:
$V_{gold}=\frac{3.81-(2650)(3.55\times 10^{-4})}{19300-2650}=1.72\times 10^{-3}m^3$
Now $m_{gold}=\rho_{gold}V_{gold}$
$\implies m_{gold}=(19300Kg/m^3)(1.72\times 10^{-4}m^3)=3.32Kg$
(b) We know that
$f_v=\frac{\frac{m_{rock}}{V_{rock}}-\rho_{granite}}{\rho_{gold}-\rho_{granite}}$
We plug in the known values to obtain:
$f_v=\frac{\frac{3.81Kg}{3.55\times 10^{-4}m^3}-{2650Kg/m^3}}{19300Kg/m^3-2650Kg/m^3}=0.485=48.5\%$
(c) We know that
$f_m=\frac{m_{gold}}{m_{rock}}$
We plug in the known values to obtain:
$f_m=\frac{3.32Kg}{3.81Kg}=0.871=87.1\%$
